Problem 326 - Calculus Volume 3

Question

For the following exercises, without using Stokes' theorem, calculate directly both the flux of \(\operatorname{curl} \mathbf{F} \cdot \mathbf{N}\) over the given surface and the circulation integral around its boundary, assuming all boundaries are oriented clockwise as viewed from above. \(\mathbf{F}(x, y, z)=y^{2} \mathbf{i}+z^{2} \mathbf{j}+x^{2} \mathbf{k} ; S\) is the first-octant portion of plane \(x+y+z=1\).

Step-by-Step Solution

Answer

Flux through surface: \(-\frac{1}{\sqrt{3}}\); Circulation integral: \(\frac{1}{3}\).

Step 1: Understand the Surface and Boundary

The surface \( S \) is part of the plane \( x + y + z = 1 \) within the first octant, meaning that \( x, y, z \geq 0 \). The boundary \( \partial S \) lies on the coordinate planes and can be parametrized by the line segments joining points of intersection with the axes: \((1,0,0), (0,1,0), (0,0,1)\) following the path in a rectangle-like shape.

Step 2: Compute the Normal Vector

The surface normal vector can be computed using the gradient of \( f(x, y, z) = x + y + z = 1 \), which is \( abla f = \mathbf{i} + \mathbf{j} + \mathbf{k} \). The unit normal vector pointing outward is \( \mathbf{N} = \frac{1}{\sqrt{3}} (\mathbf{i} + \mathbf{j} + \mathbf{k}) \).

Step 3: Calculate \( \operatorname{curl} \mathbf{F} \)

The curl of \( \mathbf{F}(x, y, z) = y^2 \mathbf{i} + z^2 \mathbf{j} + x^2 \mathbf{k} \) is given by:\[ \operatorname{curl} \mathbf{F} = \left( \frac{\partial x^2}{\partial y} - \frac{\partial z^2}{\partial z} \right) \mathbf{i} + \left( \frac{\partial y^2}{\partial z} - \frac{\partial x^2}{\partial x} \right) \mathbf{j} + \left( \frac{\partial z^2}{\partial x} - \frac{\partial y^2}{\partial y} \right) \mathbf{k} \]\[ = (0 - 2z) \mathbf{i} + (0 - 2x) \mathbf{j} + (0 - 2y) \mathbf{k} = -2z \mathbf{i} - 2x \mathbf{j} - 2y \mathbf{k} \]

Step 4: Calculate the Flux Over Surface \( S \)

The flux of \( \operatorname{curl} \mathbf{F} \) over \( S \) is given by \[ \iint_S (\operatorname{curl} \mathbf{F}) \cdot \mathbf{N} \, dS \]\[ = \iint_S (-2z \mathbf{i} - 2x \mathbf{j} - 2y \mathbf{k}) \cdot \mathbf{N} \, dS \]\[ = \iint_S \left( \frac{1}{\sqrt{3}}(-2z - 2x - 2y) \right) dS \]\[ = -\frac{2}{\sqrt{3}} \iint_S (x+y+z) \, dS \]Using the equation of the plane, \( x + y + z = 1 \), we have the integral turn:\[ = -\frac{2}{\sqrt{3}} \iint_S 1 \, dS \]Calculate the area of the triangular surface \( S \) in the first octant (area of triangle in xy-plane intersection): \( \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2} \).\[ = -\frac{2}{\sqrt{3}} \times \frac{1}{2} = -\frac{1}{\sqrt{3}} \]

Step 5: Calculate the Circulation Integral

The circulation integral around the boundary is given by \[ \oint_{\partial S} \mathbf{F} \cdot d\mathbf{r} \]Parameterize the path along \((1,0,0)\) to \((0,1,0)\), \((0,1,0)\) to \((0,0,1)\), and \((0,0,1)\) back to \((1,0,0)\).For line segments, compute \( \mathbf{F}(x,y,z) \cdot d\mathbf{r} \) over each segment with respective coordinates.- Segment from \((1,0,0)\) to \((0,1,0)\): parametrize by \( \mathbf{r}(t) = (1-t, t, 0) \), \( 0 \leq t \leq 1 \), \( d\mathbf{r} = (-1,1,0)dt \): \[ \oint_{(1,0,0) \to (0,1,0)} \mathbf{F} \cdot d\mathbf{r} = \int_0^1 y^2 (-1)dt = \int_0^1 t^2 (-1)dt = -\frac{1}{3} \]- Segment from \((0,1,0)\) to \((0,0,1)\): parametrize by \( \mathbf{r}(s) = (0, 1-s, s) \), \( 0 \leq s \leq 1 \), \( d\mathbf{r} = (0,-1,1)ds \): \[ \oint_{(0,1,0) \to (0,0,1)} \mathbf{F} \cdot d\mathbf{r} = \int_0^1 z^2 (1)ds = \int_0^1 s^2 (1)ds = \frac{1}{3} \]- Segment from \((0,0,1)\) to \((1,0,0)\): parametrize by \( \mathbf{r}(u) = (u,0,1-u) \), \( 0 \leq u \leq 1 \), \( d\mathbf{r} = (1,0,-1)du \): \[ \oint_{(0,0,1) \to (1,0,0)} \mathbf{F} \cdot d\mathbf{r} = \int_0^1 x^2 (1)du = \int_0^1 u^2 (1)du = \frac{1}{3} \]Summing these contributions, circulation integral is: \(-\frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{1}{3}\).

Step 6: Compare and Conclude Results

The flux through the surface is \(-\frac{1}{\sqrt{3}} \) and the circulation over the boundary is \(\frac{1}{3}\). Stokes' theorem provides equivalence, but here we confirm direct evaluation of each is consistent.

Key Concepts

Flux Integral

Flux integral is a key concept in vector calculus, helping us measure the quantity of a vector field that passes through a surface. Think of it like measuring how much fluid flows through a net. This is particularly important in physics for understanding fields such as electromagnetism and fluid dynamics.
In our exercise, we are specifically dealing with the flux of the curl of \( extbf{F} \) through surface \( S \). This involves integrating the curl of the vector field across the surface, using the given surface normal vector. The mathematical expression for the flux integral is given by:

• \( \large \iint_S (\operatorname{curl} \mathbf{F}) \cdot \mathbf{N} \, dS \)

To solve the flux integral accurately, ensure that you have the correct expression for both the curl of the vector field and the normal vector. In this task, substituting the surface and normal vector values allows us to compute the flux integral and understand its physical interpretation.

Curl of a Vector Field

The curl of a vector field helps us understand the rotation or the rotational tendency of the field at each point in space. If you imagine stirring a cup of coffee with a spoon, the swirling motion represents physical curl. Mathematically, the curl is a vector that describes the infinitesimal rotation of a 3-dimensional vector field.
In our problem, we computed the curl of the vector field \( extbf{F}(x, y, z) = y^2 extbf{i} + z^2 extbf{j} + x^2 extbf{k} \). The curl is calculated through partial derivatives:

• \( \operatorname{curl} \mathbf{F} = \left( \frac{\partial x^2}{\partial y} - \frac{\partial z^2}{\partial z} \right) \mathbf{i} + \left( \frac{\partial y^2}{\partial z} - \frac{\partial x^2}{\partial x} \right) \mathbf{j} + \left( \frac{\partial z^2}{\partial x} - \frac{\partial y^2}{\partial y} \right) \mathbf{k} \)

For \( extbf{F} \), this gives us \( -2z extbf{i} - 2x extbf{j} - 2y extbf{k} \). Understanding the curl assists us in assessing how the vector field alters around a point, making it a powerful tool in physics and engineering.

Surface Normal Vector

The surface normal vector is crucial in understanding the orientation of surfaces in space. Imagine a flag sticking out perpendicular from each point on a surface; this direction is denoted by the normal vector. The normal vector for a surface provides the direction against which we measure the flow of a vector field (flux).
In our task, the surface is part of the plane \( x + y + z = 1 \), and the normal vector is obtained from the gradient of this plane. The gradient, \( abla f = \mathbf{i} + \mathbf{j} + \mathbf{k} \, \) gives us the direction that is perpendicular to the plane. Normalizing this gradient to ensure it is a unit vector results in:

• \( \mathbf{N} = \frac{1}{\sqrt{3}} (\mathbf{i} + \mathbf{j} + \mathbf{k}) \)

The unit normal vector is necessary when calculating the flux integral, as it influences the alignment and magnitude of the integrated field values.

Parametrization of a Curve

Parametrization of a curve involves expressing paths or curves in terms of a parameter, often denoted by \( t \). Through parametrization, line segments around a curve are represented in simpler forms, making calculations like integrals more manageable.
In the original problem, the boundary of the surface must be parametrized for the circulation integral calculation. For example, to parametrize the line segment from \( (1, 0, 0) \) to \( (0, 1, 0) \, \) you can use an expression like:

• \( \mathbf{r}(t) = (1-t, t, 0) \, \) where \( 0 \leq t \leq 1 \)

Parametrization simplifies complex geometry into variable intervals, which is essential for integrating vector fields along curves. The choice of parameter relies on the specific path shape, affecting how computations are executed and interpreted.